Why Adjust a Likelihood-Ratio Statistic?
For a regular nested-model test, the likelihood-ratio statistic has an asymptotic chi-square distribution under the null hypothesis. At a finite sample size, however, its null distribution can differ systematically from that limit. In many smooth, correctly specified models, the leading discrepancy acts like a small scale inflation: the statistic behaves approximately like a chi-square variable multiplied by \(1+b/n\). Dividing by this factor is the Bartlett correction.
The adjustment is not a replacement for the likelihood-ratio test or a remedy for every small-sample problem. It targets a particular order-\(n^{-1}\) error in regular likelihood inference. Its validity depends on a smooth higher-order approximation, not merely on the fact that the statistic has approximately the right mean.
Conditions and the Distributional Argument
This expansion is a higher-order result for a regular likelihood problem. A standard setting has independent observations, a fixed-dimensional parameter, a correctly specified likelihood, and a null hypothesis given by \(q\) smooth restrictions. The true parameter must be an interior point; the relevant maximum-likelihood estimators must exist and be consistent; and the Fisher information must be nonsingular. The log likelihood must have sufficiently many continuous derivatives near the true parameter, with suitable moment and tail conditions for the likelihood expansion and its remainder. The restrictions must have full rank locally. These assumptions rule out, for example, boundary parameters, nonidentifiability, and singular information.
The displayed expansion states the additional higher-order property needed for the simple Bartlett adjustment: through order \(n^{-1}\), the departure from the chi-square distribution must have the particular scale form shown. In regular fixed-dimensional likelihood models, likelihood Taylor expansions and cumulant calculations yield this form under suitable strengthened smoothness and moment conditions. The value of \(b\) depends on the model and null parameter; it is not a universal constant. In a generalized model, the calculation must use the full likelihood and its design and nuisance-parameter structure.
\(W/c\le x\) is equivalent to \(W\le cx\), so its probability is \(F_W(cx)\).
Substitute \(cx\) into the expansion: \(F_W(cx)=F_q(cx)-(b/n)cx f_q(cx)+O(n^{-2})\).
Since \(c=1+b/n\), Taylor expansion gives \(F_q(cx)=F_q(x)+(b/n)x f_q(x)+O(n^{-2})\). Also, \(cx f_q(cx)=x f_q(x)+O(n^{-1})\), so its contribution after multiplication by \(b/n\) is \((b/n)x f_q(x)+O(n^{-2})\).
The positive and negative terms of order \(n^{-1}\) cancel, leaving \(F(W/c\le x)=F_q(x)+O(n^{-2})\).
The same expansion explains the usual mean-matching interpretation. If the expansion and its remainder are sufficiently well behaved to integrate over the upper tail, then $$ E[W]=\int_0^\infty \{1-F_W(x)\}\,dx =q+\frac{b}{n}\int_0^\infty x f_q(x)\,dx+O(n^{-2}) =q\left(1+\frac{b}{n}\right)+O(n^{-2}), $$ because a \(\chi_q^2\) variable has mean \(q\). This gives a useful way to recognize or check a Bartlett coefficient. But matching the mean alone does not prove the distributional expansion: other departures from a chi-square distribution can have the same mean.
Worked Examples
Worked Example: A Normal Mean With Known Variance
Let \(X_1,\ldots,X_n\) be independent \(N(\mu,\sigma^2)\) observations, where \(\sigma^2\) is known. Test \(H_0:\mu=\mu_0\). The unrestricted estimator is \(\hat\mu=\bar X\), and under the null the restricted estimator is \(\tilde\mu=\mu_0\).
The log likelihood, up to a term not involving \(\mu\), is \(\ell(\mu)=-\sum_{i=1}^n(X_i-\mu)^2/(2\sigma^2)\). Therefore $$ W=2\{\ell(\bar X)-\ell(\mu_0)\} =\frac{1}{\sigma^2}\left\{\sum_{i=1}^n(X_i-\mu_0)^2-\sum_{i=1}^n(X_i-\bar X)^2\right\} =\frac{n(\bar X-\mu_0)^2}{\sigma^2}. $$ Under \(H_0\), \(\sqrt{n}(\bar X-\mu_0)/\sigma\sim N(0,1)\), so \(W\sim\chi_1^2\) exactly for every \(n\). There is no first-order size discrepancy: \(b=0\), and the Bartlett correction leaves \(W\) unchanged. This example illustrates that the correction is not automatically nonzero in finite samples.
Worked Example: Exponential Rate With a Point Null
Suppose \(X_1,\ldots,X_n\) are independent exponential observations with rate \(\lambda\), so their density is \(f(x;\lambda)=\lambda e^{-\lambda x}\) for \(x>0\). Test \(H_0:\lambda=\lambda_0\), with \(\lambda_0\) an interior positive value. The unrestricted maximum-likelihood estimator is \(\hat\lambda=1/\bar X\).
Writing \(S=\sum_iX_i=n\bar X\), the log likelihood is \(\ell(\lambda)=n\log\lambda-\lambda S\). Substitution of the unrestricted and null estimates gives $$ W=2\{\ell(\hat\lambda)-\ell(\lambda_0)\} =2n\{\lambda_0\bar X-1-\log(\lambda_0\bar X)\}. $$ Under \(H_0\), \(T=\lambda_0S\) has a gamma distribution with shape \(n\) and rate \(1\), and \(\lambda_0\bar X=T/n\).
To find the mean exactly, use \(E[T]=n\) and \(E[\log T]=\psi(n)\), where \(\psi(n)=\frac{d}{dn}\log\Gamma(n)\) is the digamma function. The second identity follows by differentiating the gamma integral \(\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt\) with respect to \(a\), then setting \(a=n\). Hence $$ E[W]=2n\left\{E[T/n]-1-E[\log(T/n)]\right\} =2n\{\log n-\psi(n)\}. $$ The large-\(n\) expansion \(\psi(n)=\log n-\frac{1}{2n}-\frac{1}{12n^2}+O(n^{-4})\) yields $$ E[W]=1+\frac{1}{6n}+O(n^{-3}). $$ Since \(q=1\), the first-order Bartlett coefficient is \(b=1/6\). The corrected statistic is \(W_{\mathrm B}=W/(1+1/(6n))\). The mean calculation identifies the coefficient here; the higher-order likelihood expansion supplies the distributional justification for using the Bartlett correction.
Worked Example: Adjusting a Five-Percent Critical Value
Continue with the exponential rate test and take \(n=10\). Its first-order correction factor is $$ c=1+\frac{1}{6n}=1+\frac{1}{60}=\frac{61}{60}\approx1.01667. $$ Suppose the uncorrected likelihood-ratio statistic is compared with the five-percent upper critical value of \(\chi_1^2\), approximately \(3.841\). The corrected test rejects when $$ \frac{W}{c}>3.841, \qquad\text{equivalently}\qquad W>3.841\left(\frac{61}{60}\right)\approx3.905. $$ Thus the adjustment raises the rejection threshold for the uncorrected statistic. This direction is consistent with its mean being inflated above \(1\). The calculation is a first-order correction, not an exact finite-sample critical value.
Using the Correction Carefully
In applications, the coefficient \(b\) is obtained from higher-order likelihood calculations for the particular model, often through cumulants of likelihood derivatives. In a model with nuisance parameters, those calculations must account for estimation of the nuisance parameters as well as the tested parameters. The exponential example is unusually simple because its null statistic has an exact gamma-based representation and its mean gives the coefficient transparently.
The correction matters because even a modest scale inflation can affect rejection rates when the nominal significance level is small. It can improve calibration without changing the restriction being tested or the basic likelihood-ratio construction. It also makes clear that the ordinary chi-square approximation and its first finite-sample refinement are distinct statements: the first is a leading asymptotic limit, while the second needs additional regularity and model-specific calculation.
Common mistake: “The observed likelihood-ratio statistic has a mean close to \(q(1+b/n)\), so dividing by \(1+b/n\) must remove every order-\(n^{-1}\) error.” Mean matching alone does not establish this. The correction improves the null distribution at first order only when the likelihood expansion has the Bartlett form, in which the entire leading CDF error is the scale term \(-(b/n)x f_q(x)\). A different order-\(n^{-1}\) shape error would not be removed by a scalar rescaling.
The usual correction should not be transferred mechanically to nonregular settings. Boundary null parameters, unidentified parameters, singular information, a growing number of restrictions, or a misspecified likelihood can invalidate the stated expansion. A quasi-likelihood or estimating-equation statistic also does not automatically inherit a likelihood Bartlett factor; the likelihood-based regularity argument is essential.
Check Your Understanding
Answer these questions using the definitions and arguments above.
- In the Bartlett expansion, what particular form must the order-\(n^{-1}\) CDF error take for a scalar rescaling to cancel it?
- Why does the expansion imply \(E[W]=q(1+b/n)+O(n^{-2})\), and why does this mean relation alone not prove the correction is valid?
- For the known-variance normal mean test, explain why the Bartlett coefficient is zero.
- For the exponential rate test, state the exact null distribution of \(T=\lambda_0\sum_iX_i\) and the first-order value of \(b\).
- Name two conditions or settings that can invalidate the regular Bartlett expansion.